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During the break of a football match the coach will make 3 substitutions. If the team consists of 11 players among which there are 2 forwards, what is the probability that none of the forwards will be substituted?

The probablity is: How many ways that only players among the 9 are picked for the 3 spots _______________________________________________________________(Divided By) How many ways that any of the 11 players may be picked for 3 spots

So prob is: 9!/ (6!3!) - ways to rearrange 9 players into 3 spots, and order doesn't matter ___________________________________________________________________(Divided By) 11! (8!3!) - ways to rearrange 11 players into 3 spots, and order doesn't matter

Since 2 are forward players , the probability of the first substitution not being one of them will be (11-2)/11. Now there are only 10 players left of which 8 are eligible to be substituted .. so the probability of that will be 8/10

and similarly the last substitution will be - 7/9

therefore the total probability will be .. 9/11 * 8/10 * 7/9 = 28/55

_______________________________________________ Please advise if this is the correct approach .

During the break of a football match the coach will make 3 substitutions. If the team consists of 11 players among which there are 2 forwards, what is the probability that none of the forwards will be substituted?

My 5-step method for probability of multiple events:

1) Lay out the number of events (3 substitutions): _ _ _

2) Label the events with one specific example of the desired outcome (3 non-forwards): _ _ _ NF NF NF

3) Assign the relevant probability of each event and multiply across (start with 11 players and 9 non-forwards): 9/11 8/10 7/9 _ _ _ NF NF NF product = 28/55

4) Determine the number of ways in which we can have the desired outcome (only one way to have three NF) = 1

5) Multiply the result of step 3 by the result of step 4 (ie add the probabilities of each combination of desired outcome): 28/55 X 1 = 28/55

The ways to substitute non-forwards=9C3=9!/(9-3)!=9!/6!=9*8*7*6!/6!=9*8*7 The ways to substitute any of the players (or total ways of substituting players)=11C3=11!/(11-3)!=11!/8!=11*10*9*8!/8!=11*10*9

Now Probability for substituting non-forwards=The ways to substitute non-forwards/The ways to substitute any of the players ==> Probability for substituting non-forwards=(9*8*7)/(11*10*9)=(8*7)/(11*10)=56/110=28/55 So the answer is C
_________________

"Don't be afraid of the space between your dreams and reality. If you can dream it, you can make it so." Target=780 http://challengemba.blogspot.com Kudos??

Since 2 are forward players , the probability of the first substitution not being one of them will be (11-2)/11. Now there are only 10 players left of which 8 are eligible to be substituted .. so the probability of that will be 8/10

and similarly the last substitution will be - 7/9

therefore the total probability will be .. 9/11 * 8/10 * 7/9 = 28/55

_______________________________________________ Please advise if this is the correct approach .

Your method is correct. You can also solve by combination as given by few other people.

This question is awkward. If the "team" is composed of eleven players then that should include the substitutes so the actual probability of of no forwards being selected on the first substitution is 11-3-2=6. Or, 6/11. If you assume the substitutes that have entered the match cannot be taken out then the probability should look like 6/11*5/10*4/9

Out of 11 players 3 can be substituted in 11C3 ways.

Since 2 players are forward players, and these two are not suppose to be replaced. Now as per given condition, we have to substitute 3 players out of 9. SO number of ways are 9C3.

Probability = No of favorable ways/Total number of ways = 9C3/11C3

My approach: probability of selecting both forwards(a) = 9C1/11C3 probability of selection any one forward(b) = (9C2x2C1)/11C3 probability of selecting no forward = 1-(a+b)

probability of a forward being first sub: 2/11. p( not forward)= 9/11 probability of a forward being second sub: 2/10 p(not forward)= 8/10 prob of a frward being third sub:2/9 p(not forward)= 7/9

prob of no forwar din all 3 subs= 9/11*8/10*7/9 = 28/55

No forward is substituted -> choose 3 persons to substitute from the remaining 9 players -> P(No forward is substituted) = No of ways to substitute from 9 players/Total no of ways = 9C3/11C3=28/55=> C